If f(x) is continuous at x=0, then limx→0−f(x)=limx→0+f(x)=f(0)
LHL=limh→0[(0−h)sin(10−h)]=limh→0hsin1h=0×numberbet.0and1=0
RHL=limh→0hsin1h=0×number=0
Hence LHL=RHL=f(0). The given function is continuous at x=0
.
f′(x)=limh→0hsin1h−f(0)h=limh→0sin1h−0=limh→0sin1h
Since this limit does not exist, the given function is not differentiable at x=0